The proportion of TikTok users ages $$ 40-49 $$ is estimated to be $$ 20.3 \% $$. In a randomly selected sample of 147 TikTok users, find the probability that you observe $$ 20 \% $$ or less users aged $$ 40-49 $$. You may use the Normal Distribution tool. Mean should not be rounded, standard deviation may be rounded to four decimal places. Round proportions to four decimal places or percentages to two decimal places. If entering a percentage, include the \% symbol.

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We let $$ X $$ be the number of TikTok users in the sample of 147 who are ages $$40-49$$. We are given that the true proportion is  
$$
p = 0.203,
$$
so $$ X $$ follows a binomial distribution with parameters $$ n = 147 $$ and $$ p = 0.203 $$.

Because the sample size is large we approximate the binomial with a normal distribution. The mean and variance for $$ X $$ are

$$
\mu = np = 147 \times 0.203 = 29.841,
$$
$$
\sigma^2 = np(1-p) = 147 \times 0.203 \times 0.797.
$$

Calculating the variance:
$$
147 \times 0.203 \approx 29.841,
$$
$$
29.841 \times 0.797 \approx 23.756.
$$
Thus, the standard deviation is
$$
\sigma = \sqrt{23.756} \approx 4.8750.
$$

We are to find the probability that $$ 20\% $$ or less of the users are aged $$40-49$$. In the sample this corresponds to at most
$$
0.20 \times 147 = 29.4
$$
users. Since $$ X $$ is discrete, the event “20% or less” means $$ X \leq 29 $$. To use the normal approximation, we apply the continuity correction and use the value $$ 29.5 $$.

The $$ z $$-score is then
$$
z = \frac{29.5 - \mu}{\sigma} = \frac{29.5 - 29.841}{4.8750} \approx \frac{-0.341}{4.8750} \approx -0.0699.
$$

Using the standard normal distribution, the probability is
$$
P(X \leq 29) \approx P(Z \leq -0.0699) \approx 0.4721.
$$

Thus, the probability that $$20\%$$ or less of the 147 users are aged $$40-49$$ is approximately $$0.4721$$ (or $$47.21\%$$).
The probability that 20% or fewer of the 147 TikTok users are aged 40-49 is approximately 47.21%.

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